Introduction to the Leibniz Rule

In calculus, there are different rules of differentiation, used according to the nature of the given function. For example, we use the power rule if there is a polynomial function with some exponent. But the question is how we will find the derivative of a product of two functions when they are n-times differentiable. General Leibniz rule is used for this purpose. Let us understand the use of the Leibniz rule to find derivatives. 

Understanding of the Leibniz Rule

The Leibniz rule is a rule of differentiation which is used when two functions are differentiable n times. This rule generalises the product rule of derivatives. Since the product rule is used to calculate derivative of a product of two functions at a time. Similarly, the Leibniz rule is used to find derivatives of a product of two functions when they are n times differentiable.

The Leibniz rule states that if two functions f(x) and g(x) are individually n times differentiable, then their product f(x)g(x) is also differentiable n times. This rule is represented by using a summation formula which can also be generalised to find the second derivative of the product of two functions. 

Leibniz Rule Formula

The Leibniz rule is named after a mathematician Gottfried Wilhelm Leibniz that generalised the product rule for two functions multiplied together. If two function f(x) and g(x) are n-time differentiable functions, then the Leibniz rule for the product of both functions will be represented as;

$\frac{d^n}{dx^n}[f(x)g(x)]=\sum_{k=0}^nf^{(n-k)}g^{(k)}$

Where, 

$(n k)=\frac{n!}{k!(n-k)!}$ is the binomial coefficient and f’ denotes the derivative of the function. This rule can be proved by using the product rule of differentiation. For example, if n=k=1 then the Leibniz rule will be converted to, 

$\frac{d}{dx}[f(x)g(x)]=f(x)g'x+f'(x)g(x)$

Moreover, we can use the Leibniz rule to find second derivative of a product of two functions which is,

$\frac{d^2}{dx^2}[f(x)g(x)]=g(x).f''(x)+2f'(x)g'(x)+f(x).g''(x)$

Where, 

• f’(x) and f’’(x) is the first and second derivative of the function f(x).
• g’(x) and g’’(x) is the first and second derivative of the function g(x).

How to apply the general Leibniz Rule?

Like the product rule of derivative, we can apply the Leibniz rule by using the following steps.

1. Identify the first and second function in the given expression.
2. Named the two functions as f(x) and g(x).
3. Identify how many times you want to calculate the derivative of the product f(x)g(x). 
4. Use the Leibniz rule formula and substitute the value of n and r. If n = 1, it means that you have to calculate the first derivative.
5. Find the derivatives involved in the Leibniz rule. 
6. Simplify if needed.

Let us understand how to apply the Leibniz rule in the following example.

General Leibniz Rule Example 1

Find the derivative of the product of f(x) = x4 and g(x) = ln x by using Leibniz rule.

Given that,

$f(x) = x^4$

$g(x) = \ln x$

The Leibniz rule of differentiation is,

$\frac{d^n}{dx^n}[f(x)g(x)]=\sum_{k=0}^nf^{(n-k)}g^{(k)}$

Since we have to calculate first derivative, therefore the Leibniz rule reduced to, 

$\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$

Since the derivative of x4 is 4x3 and the derivative of ln x is 1/x. so, 

$\frac{d}{dx}[f(x)g(x)] = 4x^3 \ln x + x^4\left(\frac{1}{x}\right)$

$d/dx[f(x)g(x)] = 4x^3 \ln x + x^3$

Or,

$\frac{d}{dx}[f(x)g(x)] = x^3 (4 \ln x + 1)$

Hence the derivative of given functions by using the Leibniz rule is x3 (4ln x + 1).

General Leibniz Rule Example 2

Find the second derivative of the product of f(x) = x4 and g(x) = cos x by using Leibniz rule.

Given that,

$f(x) = x^4$

$g(x) = \cos x$

The Leibniz rule of differentiation is,

$\frac{d^n}{dx^n}[f(x)g(x)]=\sum_{k=0}^nf^{(n-k)}g^{(k)}$

Since we have to calculate second derivative, therefore the Leibniz rule reduced to, 

$\frac{d^2}{dx^2}[f(x)g(x)]=g(x).f''(x)+2f'(x)g'(x)+f(x).g''(x)$

Since the first derivative of x4 is 4x3 and the derivative of cos x is -sin x. The second derivatives of both functions are

$f''(x) = 12x^2$

$g''(x)=-\cos x$

Now by using Leibniz rule

$\frac{d^2}{dx^2}[f(x)g(x)] = 12x^2\cos x + 2(4x^3)(-\sin x)+x^4.(-\cos x)$

$\frac{d^2}{dx^2}[f(x)g(x)] = 12x^2\cos x – 8x^3\sin x – x^4\cos x$

Or,

$\frac{d^2}{dx^2}[f(x)g(x)] = x^2 (2\cos x – 8x\sin x – x^2\cos x)$

Hence the derivative of given functions by using Leibniz rule is x2 (2cos x – 8xsin x – x2cos x)

Comparison between Leibniz rule and Product rule 

The comparison between the product rule and Leibniz rule can be easily analysed using the following difference table.

Conclusion

In calculus, the Leibniz rule is a rule of differentiation. It is used when two functions are n-times differentiable. It means that we can find their derivatives up to nth order. The General Leibniz rule states that if two functions are n-times differentiable, then their product is also differentiable n-time. We can conclude that, by using the Leibniz rule formula, we can find higher-order derivatives of two functions multiplied together.